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The Stone Representation theorem states that every Boolean algebra is isomorphic to a field of sets. That is, a Boolean algebra whose elements are sets, and sums, products, negation are union, intersection and complements respectively.

But the theorem doesn't say that we preserve any more than the basic Boolean structure. One of my undergrad teachers does in fact work a lot around Boolean algebras. So once I asked him,

If we have a $\sigma$-algebra (or a complete Boolean algebra) is the field of sets from Stone's theorem closed under countable unions (or arbitrary unions)?

His answer was no. But he never gave me an example.

Question. Was my professor right (probably yes), and if so, what is an example of a $\sigma$-algebra (or even a complete Boolean algebra) which is not isomorphic to any $\sigma$-field of sets (or a field of set closed under arbitrary unions)?

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You probably already know examples of this without realising it!

Let $\kappa$ be an uncountable cardinal and let $P$ be the poset of finite partial injections $\kappa \to \omega$, ordered by $\supseteq$. For each $i < \kappa$ and each $n < \omega$, let $p_{i,n} : \kappa \rightharpoonup \omega$ be the partial map defined only on $i$ with value $n$. Here, an ideal of $P$ is a downward-closed subset $I \subseteq P$ with the following property:

  • Let $f \in P$, let $i < \kappa$, and suppose $i \notin \operatorname{dom} f$. For each $n < \omega$, let $f_n$ be $f \cup p_{i,n}$ (i.e. $f \land p_{i,n}$ w.r.t. the ordering of $P$). If $f_n \in I$ for all $n < \omega$, then $f \in I$.

Let $\Omega$ be the poset of ideals of $P$. Clearly, $\Omega$ is closed under arbitrary intersections, so $\Omega$ is a complete lattice. (In fact, $\Omega$ is a complete Heyting algebra, but we will not need this fact.) It is easy to see that the set $${\downarrow} (f) = \{ p \in P : p \supseteq f \}$$ is an ideal of $P$, so we get an order-embedding ${\downarrow} : P \to \Omega$. Note that it preserves all meets that exist in $P$.

I claim that there do not exist $\sigma$-prime filters of $\Omega$, i.e. upward-closed subsets $F \subset \Omega$ such that:

  • The bottom element of $\Omega$ is not in $F$.
  • The top element of $\Omega$ is in $F$, and $F$ is closed under finite meets.
  • If $\left( \bigvee_{n < \omega} I_n \right) \in F$, then for some $n < \omega$, $I_n \in F$.

Indeed, suppose $F$ is a $\sigma$-prime filter of $\Omega$. Then $\bigvee_{n < \omega} {\downarrow} (p_{i,n})$ is the top element of $\Omega$; hence, for some $f (i) < \omega$, ${\downarrow} (p_{i, f(i)}) \in F$. (In fact, $f (i)$ is uniquely determined by $F$.) Moreover, ${\downarrow} (p_{i, f(i)}) \cap {\downarrow} (p_{j, f(j)}) = {\downarrow} (p_{i, f(i)} \land p_{j, f(j)}) \in F$, so $i \ne j$ implies $f (i) \ne f (j)$. So we have an injective total map $f : \kappa \to \omega$ – a contradiction!

It follows from the claim that every monotone map $\Omega \to 2$ that preserves countable joins must be a constant map. In particular, there cannot exist an order-embedding $\Omega \to 2^X$ that preserves countable joins. Although $\Omega$ is not a boolean algebra, there does exist a non-trivial complete boolean algebra $B$ and an order-embedding $\Omega \to B$ which preserves arbitrary joins and finite meets. (This is a theorem of Barr, and it uses the fact that $\Omega$ is a complete Heyting algebra.) Then $B$ will be a $\sigma$-algebra not isomorphic to any $\sigma$-algebra of sets.

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  • $\begingroup$ In the first bullet (the property of ideals) do you mean $f\in P$ rather that in $\Omega$? $\endgroup$ – Asaf Karagila Apr 2 '14 at 13:47
  • $\begingroup$ It seems to me that there is something hiding here which is really a statement about homogeneous forcings (in this case, that are not $\sigma$-closed, but in the general case, I suppose, you can just take the least $\kappa$ that the forcing is not closed for $\kappa$ sequences and you will get a Boolean algebra which is $\kappa$-closed but not $\kappa$-algebra of sets; if I'm correct about this intuition). But I get the gist! $\endgroup$ – Asaf Karagila Apr 2 '14 at 13:58
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    $\begingroup$ I'm not sufficiently familiar with forcing to understand what you said. The idea for me is this: a complete Heyting/boolean algebra admits an order-embedding in a powerset that preserves finite meets and arbitrary joins if and only if it has enough "points", i.e. completely-prime filters; so we should be able to extract an example from any consistent theory in infinitary propositional logic that has no 2-valued models. $\endgroup$ – Zhen Lin Apr 2 '14 at 14:10
  • $\begingroup$ Well, the partial order you defined is essentially the Levy collapse of $\kappa$ (i.e. a forcing which makes $\kappa$ countable); not exactly but we can prove that this and the "standard definition" are equivalent. The argument for the non-existence of $\sigma$-prime filters resembles arguments against the closure of the forcing, and arguments of genericity. Anyway, as I wrote, I think I get the gist. Thank you! $\endgroup$ – Asaf Karagila Apr 2 '14 at 14:19
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The question in the quote box is different from the question at the end of your post - it's possible to have a $\sigma$-algebra/complete Boolean algebra which is isomorphic to an algebra of sets, but for which the field of sets given by Stone's theorem is not closed under infinite unions.

In fact, the easiest example works: Consider $B = \mathcal{P}(\omega)$ as either a $\sigma$-algebra or a complete Boolean algebra. Now $[\{n\}] = \{U\in S(B)\mid \{n\}\in U\} = \{\text{the principal ultrafilter generated by }n\}$ is a clopen set for each $n\in \omega$, but $\bigcup_{n\in\omega}[\{n\}] = \{U\in S(B)\mid U\text{ is principal}\}$ is not equal to $[X]$ for any $X\subseteq \omega$.

Of course, $B$ is a field of sets. Zhen Lin's answer addresses the question you actually care about.

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Your professor is right.

As you said, Stone's representation theorem asserts that every Boolean algebra is isomorphic to a field of set. However, not every Boolean $\sigma$-algebra is $\sigma$-isomorphic to a $\sigma$-field of sets.

A typical example is the quotient of the algebra of Lebesgue measurable subsets of the unit interval by the ideal of sets of measure zero. It is a Boolean $\sigma$-algebra which is not $\sigma$-isomorphic to any $\sigma$-field of sets (a result proved, if I am not mistaken, by Marczewski (1946): Mesure dans les corps de Boole).

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  • $\begingroup$ Thanks for the reference. I had no doubts that my professor was right. His research field is effectively Boolean algebras. $\endgroup$ – Asaf Karagila May 13 '14 at 15:38

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